Tim spends 500 minutes at the school.
Step-by-step explanation:
Given,
Time for lunch = 45 minutes
Time for study hall = 45 minutes
Total = 45+45 = 90 minutes
This represents 18% of the total minutes Tim spend at school.
Let,
x be the total minutes Tim spend at school.
18% of x = 90 minutes
Dividing both sides by 0.18
Tim spends 500 minutes at the school.
Keywords: percentage, division
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We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.
First of all, we will find z-score corresponding to 38 and 56.
Now we will find z-score corresponding to 56.
We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is .
We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.
We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.
Therefore, approximately of lightbulb replacement requests numbering between 38 and 56.